Theory and Modern Applications

# Periodic orbits to Kaplan-Yorke like differential delay equations with two lags of ratio $$(2k-1)/2$$

## Abstract

In this paper, we study the periodic solutions to a type of differential delay equations with two lags of ratio $$(2k-1)/2$$ in the form

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(2k-1) \bigr) \bigr),\quad k\geq2.$$

The 4k-periodic solutions are obtained by using the variational method and the method of Kaplan-Yorke coupling system. This is a new type of differential-delay equations compared with all previous researches since the ratio of two lags is not an integer. Three functionals are constructed for a discussion on critical points. An example is given to demonstrate our main results.

## 1 Introduction

The differential delay equations have useful applications in various fields such as age-structured population growth, control theory, and any models involving responses with nonzero delays [15].

Given $$f\in C^{0}(R^{+},R^{-})$$ with $$f(-x)=-f(x),xf(x)>0,x\neq0$$. Kaplan and Yorke [6] studied the existence of 4-periodic and 6-periodic solutions to the differential delay equations

$$x'(t)=-f \bigl(x(t-1) \bigr)$$
(1.1)

and

$$x'(t)=-f \bigl(x(t-1) \bigr)-f \bigl(x(t-2) \bigr),$$
(1.2)

respectively. The method they applied is transforming the two equations into adequate ordinary differential equations by regarding the retarded functions $$x(t-1)$$ and $$x(t-2)$$ as independent variables. They guessed that the existence of $$2(n+1)$$-periodic solution to the equation

$$x'(t)=-\sum_{i=1}^{n}f \bigl(x(t-i) \bigr)$$
(1.3)

could be studied under the restriction

$$x \bigl(t-(n+1) \bigr)=-x(t),$$

which was proved by Nussbaum [7] in 1978 by use of a fixed point theorem on cones.

After then, a lot of papers [821] discussed the existence and multiplicity of $$2(n+1)$$-periodic solutions to equation (1.3) and its extension

$$x'(t)=-\sum_{i=1}^{n} \operatorname{grad} F \bigl(x(t-i) \bigr),$$
(1.4)

where $$F\in C^{1}(R^{N},R),F(-x)=F(x),F(0)=0$$.

Recently, Zhang and Ge [22] studied the multiplicity of 2n-periodic solutions to a type of differential delay equations of the form

$$x'(t)=-f \bigl(x(t-1) \bigr)-f \bigl(x(t-n) \bigr),\quad n \geq2,$$
(1.5)

and obtained new results.

In this paper, we study the periodic orbits to a type of differential delay equations with two lags of ratio $$(2k-1)/2$$ in the form

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(2k-1) \bigr) \bigr),\quad k \geq2,$$
(1.6)

which is different from (1.3) and can be regarded as a new extension of (1.2). The method applied in this paper is the variational approach in the critical point theory [23, 24].

Since the equation

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x(t-2k) \bigr),\quad k\geq2,$$

can be changed into the form of equation (1.5) by the transformation

$$t=2s,\qquad x(t)=x(2s)=y(s),\qquad\widehat{f}(y)=2f(x),$$

this paper completes the research of the equations in the form

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x(t-n) \bigr),\quad n \geq3.$$
(1.7)

In fact, it follows from

$$y'(s)=2x'(t)=-2f \bigl(x(t-2) \bigr)-2f \bigl(x(t-2k) \bigr)=-2f \bigl(x \bigl(2(s-1) \bigr) \bigr)-2f \bigl(x \bigl(2(s-k) \bigr) \bigr)$$

that

$$y'(s)=-\widehat{f} \bigl(y(s-1) \bigr)-\widehat{f} \bigl(y(s-k) \bigr)$$

for $$\widehat{f}=2f$$, which is much the same as equation (1.5).

We suppose that

$$f\in C^{0}(R,R),\quad f(-x)=-f(x),$$
(1.8)

and there are $$\alpha,\beta\in R$$ such that

$$\lim_{x \to0} \frac{f(x)}{x}= \alpha,\qquad\lim _{x \to\infty} \frac{f(x)}{x}= \beta.$$
(1.9)

Let $$F(x)=\int_{0}^{x}f(s)\,ds$$. Then $$F(-x)=F(x)$$ and $$F(0)=0$$. For convenience, we make the following assumptions.

(S1):

f satisfies (1.8) and (1.9),

(S2):

there exist $$M>0$$ and a function $$r\in C^{0}(R^{+},R^{+})$$ satisfying $$r(s)\rightarrow\infty$$, $$r(s)\rightarrow0$$ as $$s\rightarrow \infty$$ such that

$$\biggl|F(x)-\frac{1}{2}\beta x^{2}\biggr|>r\bigl(|x|\bigr)-M,$$
($$\mathrm{S}_{3}^{\pm}$$):

$$\pm[F(x)-\frac{1}{2}\beta x^{2}]>0, |x|\rightarrow\infty$$,

($$\mathrm{S}_{4}^{\pm}$$):

$$\pm[F(x)-\frac{1}{2}\alpha x^{2}]>0, 0<|x|\ll1$$.

In this paper, we need the following lemma.

Let X be a Hilbert space, $$L:X\rightarrow X$$ be a linear operator, and $$\Phi:X\rightarrow R$$ be a differentiable functional.

### Lemma 1.1

([24], Theorem 2.4; [8], Lemma 2.4)

Assume that there are two closed $$s^{1}$$-invariant linear subspaces $$X^{+}$$ and $$X^{-}$$ and $$r>0$$ such that

1. (a)

$$X^{+}\cup X^{-}$$ is closed and of finite codimensions in X,

2. (b)

$$\widehat{L}(X^{-})\subset X^{-},\widehat {L}=L+P^{-1}A_{0}$$ or $$\widehat{L}=L+P^{-1}A_{\infty}$$,

3. (c)

there exists $$c_{0}\in R$$ such that

$$\inf_{x\in X^{+}}\Phi(x)\geq c_{0},$$
4. (d)

there is $$c_{\infty}\in R$$ such that $$\Phi(x)\leq c_{\infty}<\Phi(0)=0, \forall x\in X^{-}\cap S_{r}=\{x\in X^{-}:\|x\|=r\}$$,

5. (e)

Φ satisfies $$(P.S)_{c}$$-condition, $$c_{0}< c< c_{\infty}$$. Then Φ has at least $$\frac{1}{2}[\dim(X^{+}\cap X^{-})-\operatorname{co\,dim} _{X}(X^{+}\cup X^{-})]$$ generally different critical orbits in $$\Phi ^{-1}([c_{0},c_{\infty}])$$ if

$$\bigl[\dim\bigl(X^{+}\cap X^{-}\bigr)-\operatorname{co \,dim}_{X}\bigl(X^{+}\cup X^{-}\bigr) \bigr]>0.$$

### Remark 1.1

We may use $$(P.S)$$-condition to replace condition (e) in Lemma 1.1 since $$(P.S)$$-condition implies that $$(P.S)_{c}$$-condition holds for each $$c\in R$$.

In order to construct adequate functional whose critical points are the solutions of equation (1.6), we need to distinguish our problem into three cases:

\begin{aligned}& k=3l+2, \qquad k=3l+3, \end{aligned}

and

$$k=3l+4.$$

Then we construct the corresponding three functionals.

## 2 Space X, functional Φ, and its differential $$\Phi'$$

### 2.1 4k-Periodic orbits to equation (1.6) when $$k=3l+2$$

We are concerned at the 4k-periodic solutions to (1.6) and suppose that

$$x \bigl(t-(6l+4) \bigr)=-x(t),\quad l\geq0.$$
(2.1)

We transform (1.6) into

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+3) \bigr) \bigr).$$
(2.2)

Let

\begin{aligned}& \begin{aligned}[b] \widehat{X}&= \bigl\{ x\in C_{T}:x \bigl(t-(6l+4) \bigr)=-x(t) \bigr\} \\ &= \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr):a_{i},b_{i}\in R \Biggr\} , \end{aligned} \\& \begin{aligned}[b] X={}&cl \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin \frac{(2i+1)\pi t}{6l+4} \biggr):a_{i},b_{i}\in R, \\ &{}\sum_{i=0}^{\infty}(2i+1) \bigl(a_{i}^{2}+b_{i}^{2} \bigr)< \infty \Biggr\} , \end{aligned} \end{aligned}

and define $$P:X\rightarrow L^{2}$$ by

\begin{aligned} Px(t) =&P \Biggl(\sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr) \Biggr) \\ =& \sum_{i=0}^{\infty}(2i+1) \biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr). \end{aligned}

Then the inverse $$P^{-1}$$ of P exists. For $$x\in X$$, define

\begin{aligned}& \langle x,y\rangle= \int_{0}^{12l+8} \bigl(Px(t),y(t) \bigr)\,dt,\quad\|x\|= \sqrt{ \langle x,x\rangle}, \\& \langle x,y\rangle_{2}= \int_{0}^{12l+8} \bigl(x(t),y(t) \bigr)\,dt,\quad\|x \|_{2}=\sqrt {\langle x,x\rangle_{2}}. \end{aligned}

Therefore, $$(X,\|\cdot\|)$$ is an $$H^{\frac{1}{2}}$$ space.

Define the functional $$\Phi:X\rightarrow R$$ by

\begin{aligned} \Phi(x)&=\frac{1}{2}\langle Lx,x\rangle+ \int_{0}^{4k}F \bigl(x(t) \bigr)\,dt \\ &=\frac{1}{2}\langle Lx,x\rangle+ \int _{0}^{12l+8}F \bigl(x(t) \bigr)\,dt, \end{aligned}
(2.3)

where

$$Lx=\frac{1}{2}P^{-1} \Biggl[\sum_{i=0}^{2l+1}x'(t-3i)- \sum_{i=0}^{2l}x'(t-3i-1)-\sum _{i=0}^{2l}x'(t-3i-2) \Biggr].$$
(2.4)

Let $$k_{1}= [\frac{2k-2}{3} ]= [\frac{6l+2}{3} ]=2l, k_{2}=k-1=3l+1$$, and

$$X(i)= \biggl\{ x(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin \frac{(2i+1)\pi t}{6l+4}:a_{i},b_{i}\in R \biggr\} .$$

We have

\begin{aligned} X={}&\sum_{m=0}^{\infty} \Biggl[ \sum_{i=0}^{k_{1}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr) \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr) \Biggr]. \end{aligned}
(2.5)

For each

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr)\in X,$$

define $$\Omega:X\rightarrow X$$ by

$$(\Omega x) (t)=\sum_{i=0}^{\infty} \biggl(b_{i}\cos\frac{(2i+1)\pi t}{6l+4}-a_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr)\frac{\sin\frac{(2i+1)\pi }{12l+8}}{\sin\frac{(2i+1)3\pi}{12l+8}}.$$

If $$x_{i}(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4}\in X(i), i\in N$$, then we have

$$Lx=-\frac{\pi}{12l+8} \Biggl(\Omega x+\sum_{i=0}^{\infty}x_{i} \frac{\cos\frac {(2i+1)\pi}{12l+8}}{\sin\frac{(2i+1)3\pi}{12l+8}} \Biggr).$$
(2.6)

Obviously, $$L|_{X(i)}:X(i)\rightarrow X(i)$$ is invertible.

By the theorem of Mawhin and Willem [25], Theorem 1.4, the functional Φ is differentiable, and its differential is

$$\Phi'(x)=Lx+K(x),$$
(2.7)

where $$K(x)=P^{-1}f(x)$$. It is easy to prove that $$K:(X,\|x\| ^{2})\rightarrow(X,\|x\|_{2}^{2})$$ is compact.

It is easy to see that $$\langle\Omega x,x\rangle=0$$. Therefore, from (2.6) we have that if

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin \frac{(2i+1)\pi t}{6l+4} \biggr),$$

then

\begin{aligned} &\langle Lx,x\rangle \\ &\quad=-\sum_{i=0}^{\infty}\frac{(2i+1)\pi }{2} \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \\ &\quad=\sum_{m=0}^{\infty} \Biggl[-\sum _{i=0}^{2l}\frac{(4m(3l+2)+2i+1)\pi }{2} \bigl(a_{2m(3l+2)+i}^{2} +b_{2m(3l+2)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \\ &\qquad{}+\sum_{i=0}^{2l}\frac{(4(m+1)(3l+2)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+2)-i-1}^{2} +b_{2(m+1)(3l+2)-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \\ &\qquad{}-\sum_{i=2l+1}^{3l+1}\frac{(4m(3l+2)+2i+1)\pi}{2} \bigl(a_{2m(3l+2)+i}^{2} +b_{2m(3l+2)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \\ &\qquad{}+\sum_{i=2l+1}^{3l+1}\frac{(4(m+1)(3l+2)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+2)-i-1}^{2} +b_{2(m+1)(3l+2)-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \Biggr]. \end{aligned}

On the other hand,

\begin{aligned} \bigl\langle P^{-1}\beta x,x \bigr\rangle =&\sum _{i=0}^{\infty}(6l+4)\beta \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \\ =&\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l}(6l+4)\beta \bigl(a_{2m(3l+2)+i}^{2}+b_{2m(3l+2)+i}^{2} \bigr) \\ &{}+\sum_{i=0}^{2l}(6l+4)\beta \bigl(a_{2m(3l+2)+2(3l+2)-i-1}^{2}+b_{2m(3l+2)+2(3l+2)-i-1}^{2} \bigr) \\ &{}+\sum_{i=2l+1}^{3l+1}(6l+4)\beta \bigl(a_{2m(3l+2)+i}^{2}+b_{2m(3l+2)+i}^{2} \bigr) \\ &{}+\sum_{i=2l+1}^{3l+1}(6l+4)\beta \bigl(a_{2m(3l+2)+2(3l+2)-i-1}^{2}+b_{2m(3l+2)+2(3l+2)-i-1}^{2} \bigr) \Biggr]. \end{aligned}

Therefore, we have

\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x \bigr\rangle \\ &\quad=(6l+4)\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l} \biggl(-\frac{(4m(3l+2)+2i+1)\pi}{12l+8} \frac{\cos \frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2m(3l+2)+i}^{2} +b_{2m(3l+2)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=0}^{2l} \biggl( \frac{(4(m+1)(3l+2)-2i-1)\pi}{12l+8}\frac{\cos \frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+2)-i-1}^{2} +b_{2(m+1)(3l+2)-i-1}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+1}^{3l+1} \biggl(- \frac{(4m(3l+2)+2i+1)\pi}{12l+8}\frac{\cos \frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}}+\beta \biggr) \bigl(a_{2m(3l+2)+i}^{2} +b_{2m(3l+2)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+1}^{3l+1} \biggl( \frac{(4(m+1)(3l+2)-2i-1)\pi}{12l+8}\frac {\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+2)-i-1}^{2} +b_{2(m+1)(3l+2)-i-1}^{2} \bigr) \Biggr]. \end{aligned}
(2.8)

### Lemma 2.1

Each critical point of the functional Φ is a $$(12l+8)$$-periodic solution of equation (1.6) satisfying (2.1).

### Proof

Let x be a critical point of the functional Φ. Then $$x(t)$$ satisfies

$$\frac{1}{2} \Biggl[\sum_{i=0}^{2l+1}x'(t-3i)- \sum_{i=0}^{2l}x'(t-3i-1)-\sum _{i=0}^{2l}x'(t-3i-2) \Biggr]+f \bigl(x(t) \bigr)=0.$$
(2.9)

Consequently,

\begin{aligned}& \frac{1}{2} \Biggl[\sum_{i=0}^{2l+1}x'(t-3i-2)- \sum_{i=1}^{2l}x'(t-3i)-\sum _{i=1}^{2l}x'(t-3i-1) \Biggr]+f \bigl(x(t-2) \bigr)=0, \end{aligned}
(2.10)
\begin{aligned}& \frac{1}{2} \Biggl[\sum_{i=0}^{2l+1}x'(t-3i+1)- \sum_{i=0}^{2l}x'(t-3i)-\sum _{i=0}^{2l}x'(t-3i-1) \Biggr]+f \bigl(x(t+1) \bigr)=0. \end{aligned}
(2.11)

Subtracting (2.10) from (2.11), we have

$$-x'(t)+f \bigl(x(t+1) \bigr)-f \bigl(x(t-2) \bigr)=0,$$

that is,

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+3) \bigr) \bigr),$$

which implies that x is a solution to (1.6). □

### 2.2 4k-Periodic orbits to equation (1.6) when $$k=3l+3$$

We are concerned at the 4k-periodic solutions to (1.6) and suppose that

$$x \bigl(t-(6l+6) \bigr)=-x(t),\quad l\geq0.$$
(2.12)

We transform (1.6) into

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+5) \bigr) \bigr).$$
(2.13)

Let

\begin{aligned}& \begin{aligned}[b] \widehat{X}&= \bigl\{ x\in C_{T}:x \bigl(t-(6l+6) \bigr)=-x(t) \bigr\} \\ &= \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr):a_{i},b_{i} \in R \Biggr\} , \end{aligned} \\& \begin{aligned}[b] X={}&cl \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr):a_{i},b_{i}\in R, \\ &{}\sum_{i=0}^{\infty}(2i+1) \bigl(a_{i}^{2}+b_{i}^{2} \bigr)< \infty \Biggr\} , \end{aligned} \end{aligned}

and define $$P:X\rightarrow L^{2}$$ by

\begin{aligned} Px(t) =&P \Biggl(\sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr) \Biggr) \\ =& \sum_{i=0}^{\infty}(2i+1) \biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr). \end{aligned}

Then the inverse $$P^{-1}$$ of P exists. For $$x\in X$$, define

\begin{aligned}& \langle x,y\rangle= \int_{0}^{12l+12} \bigl(Px(t),y(t) \bigr)\,dt,\quad\|x\|= \sqrt{ \langle x,x\rangle} , \\& \langle x,y\rangle_{2}= \int_{0}^{12l+12} \bigl(x(t),y(t) \bigr)\,dt,\quad\|x \|_{2}=\sqrt {\langle x,x\rangle_{2}} . \end{aligned}

Therefore, $$(X,\|\cdot\|)$$ is an $$H^{\frac{1}{2}}$$ space.

Define the functional $$\Phi:X\rightarrow R$$ by

$$\Phi(x)=\frac{1}{2}\langle Lx,x\rangle+ \int _{0}^{12l+12}F \bigl(x(t) \bigr)\,dt,$$
(2.14)

where

$$Lx=-\frac{1}{2}P^{-1}\sum_{i=0}^{2l+1}x'(t-3i-1).$$
(2.15)

Let $$k_{1}= [\frac{2k-2}{3} ]= [\frac{6l+4}{3} ]=2l+1,k_{2}=k-1=3l+2$$, and

$$X(i)= \biggl\{ x(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6}:a_{i},b_{i}\in R \biggr\} .$$

We have

\begin{aligned}[b] X={}&\sum_{m=0}^{\infty} \Biggl[ \sum_{i=0}^{k_{1}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr)\\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr) \Biggr]. \end{aligned}
(2.16)

Define $$\Omega:X\rightarrow X$$ by

$$(\Omega x) (t)=\sum_{i=0}^{\infty} \biggl(b_{i}\cos\frac{(2i+1)\pi t}{6l+6}-a_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr)\frac{\sin\frac{(2i+1)\pi }{12l+12}}{\sin\frac{(2i+1)3\pi}{12l+12}}.$$

Then if

$$x_{i}(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6}\in X(i),\quad i\in N,$$

then we have

$$Lx=-\frac{\pi}{12l+12} \Biggl(\Omega x+\sum_{i=0}^{\infty}x_{i} \frac{\cos \frac{(2i+1)\pi}{12l+12}}{\sin\frac{(2i+1)3\pi}{12l+12}} \Biggr).$$
(2.17)

The functional Φ is differentiable, and its differential is

$$\Phi'(x)=Lx+K(x),$$
(2.18)

where $$K(x)=P^{-1}f(x)$$. The mapping $$K:(X,\|x\|^{2})\rightarrow(X,\| x\|_{2}^{2})$$ is compact.

Therefore, from (2.17) it follows that, for each

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr),$$

we have

\begin{aligned} \langle Lx,x\rangle ={}&-\sum_{i=0}^{\infty}\frac{(2i+1)\pi}{2} \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \frac {\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \\ ={}&\sum_{m=0}^{\infty} \Biggl[-\sum _{i=0}^{2l+1}\frac{(4m(3l+3)+2i+1)\pi }{2} \bigl(a_{2m(3l+3)+i}^{2} +b_{2m(3l+3)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \\ &{}+\sum_{i=0}^{2l+1}\frac{(4(m+1)(3l+3)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+3)-i-1}^{2} +b_{2(m+1)(3l+3)-i-1}^{2} \bigr)\\ &{}\times \frac{\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \\ &{}-\sum_{i=2l+2}^{3l+2}\frac{(4m(3l+3)+2i+1)\pi}{2} \bigl(a_{2m(3l+3)+i}^{2} +b_{2m(3l+3)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \\ &{}+\sum_{i=2l+2}^{3l+2}\frac{(4(m+1)(3l+3)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+3)-i-1}^{2} +b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \\ &{}\times \frac{\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \Biggr]. \end{aligned}

On the other hand,

\begin{aligned}[b] \bigl\langle P^{-1}\beta x,x \bigr\rangle ={}&\sum _{m=0}^{\infty}(6l+6)\beta \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \\ ={}&\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l+1}(6l+6)\beta \bigl(a_{2m(3l+3)+i}^{2}+b_{2m(3l+3)+i}^{2} \bigr) \\ &{}+\sum_{i=0}^{2l+1}(6l+6)\beta \bigl(a_{2(m+1)(3l+3)-i-1}^{2}+b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \\ &{}+\sum_{i=2l+2}^{3l+2}(6l+6)\beta \bigl(a_{2m(3l+3)+i}^{2}+b_{2m(3l+3)+i}^{2} \bigr) \\ &{}+\sum_{i=2l+2}^{3l+2}(6l+6)\beta \bigl(a_{2(m+1)(3l+3)-i-1}^{2}+b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \Biggr]. \end{aligned}

Therefore, we have

\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x \bigr\rangle \\ &\quad=(6l+6)\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l+1} \biggl(-\frac{(4m(3l+3)+2i+1)\pi}{12l+12} \frac {\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2m(3l+3)+i}^{2} +b_{2m(3l+3)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=0}^{2l+1} \biggl( \frac{(4(m+1)(3l+3)-i-1)\pi}{12l+12}\frac{\cos \frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+3)-i-1}^{2} +b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+2}^{3l+2} \biggl(- \frac{(4m(3l+3)+2i+1)\pi}{12l+12}\frac {\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}}+\beta \biggr) \bigl(a_{2m(3l+3)+i}^{2} +b_{2m(3l+3)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+2}^{3l+2} \biggl( \frac{(4(m+1)(3l+3)-2i-1)\pi}{12l+12}\frac {\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+3)-i-1}^{2} +b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \Biggr]. \end{aligned}
(2.19)

### Lemma 2.2

Each critical point of the functional Φ is a $$(12l+12)$$-periodic solution of equation (1.6) satisfying (2.12).

### Proof

Let x be a critical point of the functional Φ. Then $$x(t)$$ satisfies

$$-\frac{1}{2}\sum_{i=0}^{2l+1}x'(t-3i-1)+f \bigl(x(t) \bigr)=0.$$
(2.20)

Consequently, we have

\begin{aligned}& -\frac{1}{2}\sum_{i=0}^{2l+1}x'(t-3i-3)+f \bigl(x(t-2) \bigr)=0, \end{aligned}
(2.21)
\begin{aligned}& -\frac{1}{2}\sum_{i=0}^{2l+1}x'(t-3i)+f \bigl(x(t+1) \bigr)=0. \end{aligned}
(2.22)

Subtracting (2.21) from (2.22), we have

$$-x'(t)+f \bigl(x(t+1) \bigr)-f \bigl(x(t-2) \bigr)=0,$$

that is,

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+5) \bigr) \bigr),$$

which implies that x is a solution to (1.6). □

### 2.3 4k-Periodic orbits to equation (1.6) when $$k=3l+4$$

We are concerned at the 4k-periodic solutions to (1.6) and suppose that

$$x \bigl(t-(6l+8) \bigr)=-x(t),\quad l\geq0.$$
(2.23)

We transform (1.6) into

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+7) \bigr) \bigr).$$
(2.24)

Let

\begin{aligned}& \begin{aligned}[b] \widehat{X}&= \bigl\{ x\in C_{T}:x \bigl(t-(6l+8) \bigr)=-x(t) \bigr\} \\ &= \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr):a_{i},b_{i} \in R \Biggr\} , \end{aligned} \\& \begin{aligned}[b] X={}&cl \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr):a_{i},b_{i}\in R, \\ &{}\sum_{i=0}^{\infty}(2i+1) \bigl(a_{i}^{2}+b_{i}^{2} \bigr)< \infty \Biggr\} , \end{aligned} \end{aligned}

and define $$P:X\rightarrow L^{2}$$ by

\begin{aligned} Px(t) =&P \Biggl(\sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr) \Biggr) \\ =& \sum_{i=0}^{\infty}(2i+1) \biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr). \end{aligned}

Then the inverse $$P^{-1}$$ of P exists. For $$x\in X$$, define

\begin{aligned}& \langle x,y\rangle= \int_{0}^{12l+16} \bigl(Px(t),y(t) \bigr)\,dt,\quad\|x\|= \sqrt{ \langle x,x\rangle} , \\& \langle x,y\rangle_{2}= \int_{0}^{12l+16} \bigl(x(t),y(t) \bigr)\,dt,\quad\|x \|_{2}=\sqrt {\langle x,x\rangle_{2}} . \end{aligned}

Therefore, $$(X,\|\cdot\|)$$ is an $$H^{\frac{1}{2}}$$ space.

Define the functional $$\Phi:X\rightarrow R$$ by

$$\Phi(x)=\frac{1}{2}\langle Lx,x\rangle+ \int _{0}^{12l+16}F \bigl(x(t) \bigr)\,dt,$$
(2.25)

where

$$Lx=\frac{1}{2}P^{-1} \Biggl[-\sum_{i=0}^{2l+2}x'(t-3i)- \sum_{i=0}^{2l+2}x'(t-3i-1)+\sum _{i=0}^{2l+1}x'(t-3i-2) \Biggr].$$
(2.26)

Let $$k_{1}= [\frac{2k-2}{3} ]= [\frac{6l+6}{3} ]=2l+2,k_{2}=k-1=3l+3$$, and

$$X(i)= \biggl\{ x(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8}:a_{i},b_{i}\in R \biggr\} .$$

We have

\begin{aligned} X={}&\sum_{m=0}^{\infty} \Biggl[ \sum_{i=0}^{k_{1}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr) \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr) \Biggr]. \end{aligned}
(2.27)

Define $$\Omega:X\rightarrow X$$ by

$$(\Omega x) (t)=\sum_{i=0}^{\infty} \biggl(b_{i}\cos\frac{(2i+1)\pi t}{6l+8}-a_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr)\frac{\sin\frac{(2i+1)\pi }{12l+16}}{\sin\frac{(2i+1)3\pi}{12l+16}}.$$

Then if

$$x_{i}(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8}\in X(i),\quad i\in N,$$

then we have

$$Lx=-\frac{\pi}{12l+16} \Biggl(\Omega x+\sum_{i=0}^{\infty}x_{i} \frac{\cos \frac{(2i+1)\pi}{12l+16}}{\sin\frac{(2i+1)3\pi}{12l+16}} \Biggr).$$
(2.28)

The functional Φ is differentiable, and its differential is

$$\Phi'(x)=Lx+K(x),$$
(2.29)

where $$K(x)=P^{-1}f(x)$$. The mapping $$K:(X,\|x\|^{2})\rightarrow(X,\| x\|_{2}^{2})$$ is compact.

For

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr),$$

we have

\begin{aligned} &\langle Lx,x\rangle \\ &\quad=-\sum_{i=0}^{\infty}\frac{(2i+1)\pi}{2} \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \frac {\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \\ &\quad=\sum_{m=0}^{\infty} \Biggl[-\sum _{i=0}^{2l+2}\frac{(4m(3l+4)+2i+1)\pi }{2} \bigl(a_{2m(3l+4)+i}^{2} +b_{2m(3l+4)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \\ &\qquad{}+\sum_{i=0}^{2l+2}\frac{(4(m+1)(3l+4)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+4)-i-1}^{2} +b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \\ &\qquad{}-\sum_{i=2l+3}^{3l+3}\frac{(4m(3l+4)+2i+1)\pi}{2} \bigl(a_{2m(3l+4)+i}^{2} +b_{2m(3l+4)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \\ &\qquad{}+\sum_{i=2l+3}^{3l+3}\frac{(4(m+1)(3l+4)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+4)-i-1}^{2} +b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \\ &\qquad{}\times \frac{\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \Biggr]. \end{aligned}

On the other hand,

\begin{aligned} \bigl\langle P^{-1}\beta x,x \bigr\rangle =&\sum _{m=0}^{\infty}(6l+8)\beta \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \\ =&\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l+2}(6l+8)\beta \bigl(a_{2m(3l+4)+i}^{2}+b_{2m(3l+4)+i}^{2} \bigr) \\ &{}+\sum_{i=0}^{2l+2}(6l+8)\beta \bigl(a_{2(m+1)(3l+4)-i-1}^{2}+b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \\ &{}+\sum_{i=2l+3}^{3l+3}(6l+8)\beta \bigl(a_{2m(3l+4)+i}^{2}+b_{2m(3l+4)+i}^{2} \bigr) \\ &{}+\sum_{i=2l+3}^{3l+3}(6l+8)\beta \bigl(a_{2(m+1)(3l+4)-i-1}^{2}+b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \Biggr]. \end{aligned}

Therefore, we have

\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x \bigr\rangle \\ &\quad=(6l+8)\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l+2} \biggl(-\frac{(4m(3l+4)+2i+1)\pi}{12l+16} \frac {\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2m(3l+4)+i}^{2} +b_{2m(3l+4)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=0}^{2l+2} \biggl( \frac{(4(m+1)(3l+4)-i-1)\pi}{12l+16}\frac{\cos \frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+4)-i-1}^{2} +b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+3}^{3l+3} \biggl(- \frac{(4m(3l+4)+2i+1)\pi}{12l+16}\frac {\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}}+\beta \biggr) \bigl(a_{2m(3l+4)+i}^{2} +b_{2m(3l+4)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+3}^{3l+3} \biggl( \frac{(4(m+1)(3l+4)-2i-1)\pi}{12l+16}\frac {\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+4)-i-1}^{2} +b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \Biggr]. \end{aligned}
(2.30)

### Lemma 2.3

Each critical point of the functional Φ is a $$(12l+16)$$-periodic solution of equation (1.6) satisfying (2.23).

### Proof

Let x be a critical point of the functional Φ. Then $$x(t)$$ satisfies

$$\frac{1}{2} \Biggl[-\sum_{i=0}^{2l+2}x'(t-3i)- \sum_{i=0}^{2l+2}x'(t-3i-1)+\sum _{i=0}^{2l+1}x'(t-3i-2) \Biggr]+f \bigl(x(t) \bigr)=0.$$
(2.31)

Consequently,

\begin{aligned}& \frac{1}{2} \Biggl[-\sum_{i=0}^{2l+2}x'(t-3i-2)- \sum_{i=0}^{2l+2}x'(t-3i-3)+\sum _{i=0}^{2l+1}x'(t-3i-4) \Biggr]+f \bigl(x(t-2) \bigr)=0, \end{aligned}
(2.32)
\begin{aligned}& \frac{1}{2} \Biggl[-\sum_{i=0}^{2l+2}x'(t-3i+1)- \sum_{i=0}^{2l+2}x'(t-3i)+\sum _{i=0}^{2l+1}x'(t-3i-1) \Biggr]+f \bigl(x(t+1) \bigr)=0. \end{aligned}
(2.33)

Subtracting (2.32) from (2.33), we have

$$-x'(t)+f \bigl(x(t+1) \bigr)-f \bigl(x(t-2) \bigr)=0,$$

that is,

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+7) \bigr) \bigr),$$

which implies that x is a solution to (1.6). □

## 3 Partition of space X and symbols

In fact, in Sections 2.1, 2.2, and 2.3, we could let

\begin{aligned}& k_{1}= \biggl[\frac{2k-2}{3} \biggr],\qquad k_{2}=k-1, \\& X(i)= \biggl\{ x(t)=a_{i}\cos\frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1)\pi t}{2k}:a_{i},b_{i}\in R \biggr\} , \end{aligned}

and

$$X=\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{k_{1}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr)+\sum _{i=k_{1}+1}^{k_{2}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr) \Biggr]$$

and define $$\Omega:X\rightarrow X$$ by

$$(\Omega x) (t)=\sum_{i=0}^{\infty} \biggl(b_{i}\cos\frac{(2i+1)\pi t}{2k}-a_{i}\sin \frac{(2i+1)\pi t}{2k} \biggr)\frac{\sin\frac{(2i+1)\pi }{4k}}{\sin\frac{(2i+1)3\pi}{4k}}.$$

Then if $$x_{i}(t)=a_{i}\cos\frac{(2i+1)\pi t}{2k}+b_{i}\sin\frac {(2i+1)\pi t}{2k}\in X(i), i\in N$$, then we have

$$Lx=-\frac{\pi}{4k} \Biggl(\Omega x+\sum_{i=0}^{\infty}x_{i} \frac{\cos\frac {(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} \Biggr),$$

and, for each

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos \frac{(2i+1)\pi t}{2k}+b_{i}\sin\frac{(2i+1)\pi t}{2k} \biggr)\in X,$$

we have

\begin{aligned} \langle Lx,x\rangle =&-\sum_{i=0}^{\infty} \frac{(2i+1)\pi }{2} \bigl(a_{i}^{2}+b_{i}^{2} \bigr)\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \\ =&\sum_{m=0}^{\infty} \Biggl[-\sum _{i=0}^{k_{1}}\frac{(4mk+2i+1)\pi }{2} \bigl(a_{2mk+i}^{2} +b_{2mk+i}^{2} \bigr)\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \\ &{}+\sum_{i=0}^{k_{1}}\frac{(4(m+1)k-2i-1)\pi}{2} \bigl(a_{2(m+1)k-i-1}^{2} +b_{2(m+1)k-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \\ &{}-\sum_{i=k_{1}+1}^{k_{2}}\frac{(4mk+2i+1)\pi}{2} \bigl(a_{2mk+i}^{2} +b_{2mk+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}}\frac{(4(m+1)k-2i-1)\pi}{2} \bigl(a_{2(m+1)k-i-1}^{2} +b_{2(m+1)k-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \Biggr]. \end{aligned}

On the other hand,

\begin{aligned} \bigl\langle P^{-1}\beta x,x \bigr\rangle =&\sum _{m=0}^{\infty}2k\beta \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \\ =&\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{k_{1}}2k\beta \bigl(a_{2mk+i}^{2}+b_{2mk+i}^{2} \bigr) \\ &{}+\sum_{i=0}^{k_{1}}2k\beta \bigl(a_{2(m+1)k-i-1}^{2}+b_{2(m+1)k-i-1}^{2} \bigr) \\ &{}+\sum_{i=k_{1}+1}^{k_{2}}2k\beta \bigl(a_{2mk+i}^{2}+b_{2mk+i}^{2} \bigr) \\ &{}+\sum_{i=k_{1}+1}^{k_{2}}2k\beta \bigl(a_{2(m+1)k-i-1}^{2}+b_{2(m+1)k-i-1}^{2} \bigr) \Biggr]. \end{aligned}

Therefore, we have

\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x \bigr\rangle \\ &\quad=2k\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{k_{1}} \biggl(-\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac {(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta \biggr) \bigl(a_{2mk+i}^{2} +b_{2mk+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=0}^{k_{1}} \biggl( \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac {(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta \biggr) \bigl(a_{2(m+1)k-i-1}^{2} +b_{2(m+1)k-i-1}^{2} \bigr) \\ &\qquad{}+\sum_{i=k_{1}+1}^{k_{2}} \biggl(- \frac{(4mk+2i+1)\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta \biggr) \bigl(a_{2mk+i}^{2} +b_{2mk+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=k_{1}+1}^{k_{2}} \biggl( \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta \biggr) \bigl(a_{2(m+1)k-i-1}^{2} +b_{2(m+1)k-i-1}^{2} \bigr) \Biggr]. \end{aligned}

Let

\begin{aligned} X_{\infty}^{+} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{} \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \biggr\} , \\ X_{\infty}^{-} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0 \biggr\} , \\ X_{0}^{+} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha>0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{} \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha>0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha>0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha>0 \biggr\} , \\ X_{0}^{-} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha< 0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha< 0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha< 0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha< 0 \biggr\} . \end{aligned}

On the other hand,

\begin{aligned} X_{\infty}^{0} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta=0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{} \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta=0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta=0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta=0 \biggr\} , \\ X_{0}^{0} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha=0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{} \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha=0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha=0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha=0 \biggr\} . \end{aligned}

Obviously, $$\dim X_{\infty}^{0}<\infty$$ and $$\dim X_{0}^{0}<\infty$$.

### Lemma 3.1

Under assumptions (S1) and (S2), there is $$\sigma>0$$ such that

$$\bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle >\sigma\|x \|^{2},\quad x\in X_{\infty}^{+},$$

and

$$\bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle < -\sigma\|x \|^{2},\quad x\in X_{\infty}^{-} .$$
(3.1)

### Proof

First, we have that, for $$\beta\geq0$$ and $$i\in\{0,1,\ldots ,k_{1}\}$$,

$$-\frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>-\frac{(4m_{0}^{+}(i)k+2i+1)\pi }{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0,$$

where $$m_{0}^{+}(i)=\max \{m\in N:-\frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac {(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \}$$, and

$$-\frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< -\frac{(4m_{0}^{-}(i)k+2i+1)\pi }{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0,$$

where $$m_{0}^{-}(i)=\min \{m\in N:-\frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac {(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta<0 \}$$.

In this case, we may choose

\begin{aligned} \sigma_{i} =&\min \biggl\{ -\frac{\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+ \frac{\beta}{4m_{0}^{+}(i)k+2i+1}, \\ &{}\frac{\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}-\frac{\beta}{4m_{0}^{-}(i)k+2i+1} \biggr\} >0. \end{aligned}

and let $$\sigma^{0}=\min\{\sigma_{0},\sigma_{1},\ldots,\sigma_{k_{1}}\}>0$$. Further, for $$\beta\geq0$$ and $$i\in\{k_{1}+1,\ldots,k_{2}\}$$, we have

$$\frac{(4mk+4k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>\frac{(4m_{1}^{+}(i)k+4k-2i-1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+ \beta>0,$$

where $$m_{1}^{+}(i)=\max \{m\in N:\frac{(4mk+4k-2i-1)\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \}$$, and

$$\frac{(4mk+4k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< \frac{(4m_{1}^{-}(i)k+4k-2i-1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+ \beta< 0,$$

where $$m_{1}^{-}(i)=\min \{m\in N:\frac{(4mk+4k-2i-1)\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta<0 \}$$.

In this case, we may choose

\begin{aligned} \sigma_{i} =&\min \biggl\{ \frac{\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+ \frac{\beta}{4m_{1}^{+}(i)k+4k-2i-1}, \\ &{}-\frac{\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}-\frac{\beta}{4m_{1}^{-}(i)k+4k-2i-1} \biggr\} >0, \end{aligned}

and let $$\sigma^{1}=\min\{\sigma_{k_{1}+1},\sigma_{k_{1}+2},\ldots ,\sigma_{k_{2}}\}>0$$.

Let $$\sigma=\min\{\sigma^{0},\sigma^{1}\}=\min\{\sigma_{1},\sigma _{2},\ldots,\sigma_{k_{2}}\}$$. The proof for the case $$\beta<0$$ is similar. We omit it. The inequalities in (3.1) are proved. □

### Lemma 3.2

Under conditions (S1) and (S2), the functional Φ defined by (2.3) satisfies $$(P.S)$$-condition if

$$\biggl|\Phi(x)-\frac{1}{2}\beta x^{2}\biggr|>r\bigl(|x|\bigr)-M_{0},\quad x \in R,$$

for some $$M_{0}>0$$ and some function $$r\in C^{0}(R^{+},R^{+})$$ that satisfies

$$r(s)\rightarrow\infty,\qquad r(s)/s\rightarrow0 \quad\textit{as } s\rightarrow \infty.$$
(3.2)

### Proof

Let $$\Pi,N,Z$$ be the orthogonal projections from X onto $$X_{\infty}^{+},X_{\infty}^{-},X_{\infty}^{0}$$, respectively. From the second condition in (1.9) it follows that

$$\bigl| \bigl\langle P^{-1} \bigl(f(x)-\beta x \bigr),x \bigr\rangle \bigr|< \frac{\sigma}{2}\|x\| ^{2}+M,\quad x\in X,$$
(3.3)

for some $$M>0$$.

Assume that $$\{x_{n}\}\subset X$$ is a subsequence such that $$\Phi '(x_{n})\rightarrow0$$ and $$\Phi(x_{n})$$ is bounded. Let $$w_{n}=\Pi x_{n},y_{n}=Nx_{n},z_{n}=Zx_{n}$$. Then we have

$$\Pi \bigl(L+P^{-1}\beta \bigr)= \bigl(L+P^{-1}\beta \bigr) \Pi,\qquad N \bigl(L+P^{-1}\beta \bigr)= \bigl(L+P^{-1}\beta \bigr)N.$$
(3.4)

From

$$\bigl\langle \Phi'(x_{n}),x_{n} \bigr\rangle = \bigl\langle Lx_{n}+P^{-1}f(x_{n}),x_{n} \bigr\rangle = \bigl\langle \bigl(L+P^{-1}\beta \bigr)x_{n},x_{n} \bigr\rangle + \bigl\langle P^{-1} \bigl(f(x_{n})-\beta x_{n} \bigr),x_{n} \bigr\rangle$$

and (3.4) we have

\begin{aligned} \bigl\langle \Pi\Phi'(x_{n}),x_{n} \bigr\rangle =& \bigl\langle \Pi \bigl(L+P^{-1}\beta \bigr)x_{n},x_{n} \bigr\rangle + \bigl\langle \Pi P^{-1} \bigl(f(x_{n})-\beta x_{n} \bigr),x_{n} \bigr\rangle \\ =& \bigl\langle \bigl(L+P^{-1}\beta \bigr)w_{n},w_{n} \bigr\rangle + \bigl\langle \Pi P^{-1} \bigl(f(x_{n})-\beta x_{n} \bigr),w_{n} \bigr\rangle \end{aligned}

and then, by (3.1),

$$\bigl\langle \bigl(L+P^{-1}\beta \bigr)w_{n},w_{n} \bigr\rangle + \bigl\langle \Pi P^{-1} \bigl(f(x_{n})-\beta x_{n} \bigr),w_{n} \bigr\rangle >\frac{\sigma}{2} \|w_{n}\| ^{2}-M\|w_{n}\|,$$

which, together with $$\Pi\Phi'(x_{n})\rightarrow0$$, implies the boundedness of $$w_{n}$$. Similarly, we have the boundedness of $$y_{n}$$. At the same time, (S2) yields

\begin{aligned} \Phi(x_{n}) =&\frac{1}{2} \bigl\langle \bigl(L+P^{-1} \beta \bigr)x_{n},x_{n} \bigr\rangle + \int_{0}^{4k}F(x_{n})\,dt-\frac{\beta}{2} \|x_{n}\|_{2}^{2} \\ =&\frac{1}{2} \bigl\langle \bigl(L+P^{-1}\beta \bigr)w_{n},w_{n} \bigr\rangle + \frac{1}{2} \bigl\langle \bigl(L+P^{-1}\beta \bigr)y_{n},y_{n} \bigr\rangle \\ &{}+ \int_{0}^{4k}F(x_{n})\,dt - \frac{\beta}{2} \bigl(\|w_{n}\|_{2}^{2}+ \|y_{n}\|_{2}^{2}+\|z_{n}\| _{2}^{2} \bigr). \end{aligned}

Then the boundedness of $$\Phi(x)$$ implies that $$\|z_{n}\|_{2}$$ is bounded. Consequently, $$\|z_{n}\|$$ is bounded since $$X_{\infty}^{0}$$ is finite-dimensional. Therefore, $$\|x_{n}\|$$ is bounded.

It follows from (2.7) that

\begin{aligned} (\Pi+N)\Phi'(x_{n}) =&(\Pi+N)Lx_{n}+( \Pi+N)Kx_{n}\\ =&L(w_{n}+y_{n})+(\Pi+N)Kx_{n}. \end{aligned}

From the compactness of operator K and the boundedness of $$x_{n}$$ we have that $$K(x_{n})\rightarrow u$$. Then

$$L|_{x_{\infty}^{+}+x_{\infty}^{-}}(w_{n}+y_{n})\rightarrow-(\Pi +N)u.$$
(3.5)

The finite-dimensionality of $$X_{\infty}^{0}$$ and the boundedness of $$z_{n}=Zx_{n}$$ imply $$z_{n}\rightarrow\varphi\in X_{\infty}^{0}$$. Therefore,

$$x_{n}=z_{n}+w_{n}+y_{n}\rightarrow \varphi-(L|_{x_{\infty}^{+}+x_{\infty }^{-}})^{-1} (\Pi+N)u,$$

which implies $$(P.S)$$-condition. □

### Lemma 3.3

Under conditions (S1) and (S2), the functional Φ defined by (2.14) and (2.25) satisfies $$(P.S)$$-condition if

$$\biggl|\Phi(x)-\frac{1}{2}\beta x^{2}\biggr|>r\bigl(|x|\bigr)-M_{0},\quad x \in R,$$

for some $$M_{0}>0$$ and some function $$r\in C^{0}(R^{+},R^{+})$$, which satisfies

$$r(s)\rightarrow\infty,\qquad r(s)/s\rightarrow0 \quad\textit{as } s\rightarrow \infty.$$

The proof of Lemma 3.3 is the same as that of Lemma 3.2, and we omit it.

## 4 Notation and main results of this paper

We first give some notation.

Denote

\begin{aligned}& N(\alpha)=\left \{ \textstyle\begin{array}{@{}l} -\sum_{i=0}^{k_{1}}\operatorname{card} \{m\geq0:0< \frac{(4mk+4k-2i-1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}}< -\alpha \}\\ \quad{}-\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \{m\geq0:0< \frac{(4mk+2i+1)\pi }{4k} |\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} |< -\alpha \}, \quad\alpha< 0,\\ \sum_{i=0}^{k_{1}}\operatorname{card} \{m\geq0:0< \frac{(4mk+2i+1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}}< \alpha \}\\ \quad{}+\sum_{i=k_{1}+1}^{k_{2}}\operatorname{card} \{m\geq0:0< \frac{(4mk+4k-2i-1)\pi}{4k} |\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} |< \alpha \}, \quad\alpha\geq0, \end{array}\displaystyle \right . \\& N(\beta)=\left \{ \textstyle\begin{array}{ll} -\sum_{i=0}^{k_{1}}\operatorname{card} \{m\geq0:0< \frac{(4mk+4k-2i-1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}}< -\beta \}\\ \quad{}-\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \{m\geq0:0< \frac{(4mk+2i+1)\pi }{4k} |\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} |< -\beta \}, \quad\beta< 0,\\ \sum_{i=0}^{k_{1}}\operatorname{card} \{m\geq0:0< \frac{(4mk+2i+1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}}< \beta \}\\ \quad{}+\sum_{i=k_{1}+1}^{k_{2}}\operatorname{card} \{m\geq0:0< \frac{(4mk+4k-2i-1)\pi}{4k} |\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} |< \beta \}, \quad \beta\geq0, \end{array}\displaystyle \right . \end{aligned}

and

\begin{aligned} N^{0}(\alpha_{-}) =&\sum_{i=0}^{k_{1}} \operatorname{\operatorname{card}} \biggl\{ m\geq0:0< \frac {(4mk+4k-2i-1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}=-\alpha \biggr\} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \biggl\{ m\geq0:0< \frac{(4mk+2i+1)\pi }{4k} \biggl|\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \biggr|=-\alpha \biggr\} , \quad \alpha< 0, \\ N^{0}(\alpha_{+}) =&\sum_{i=0}^{k_{1}} \operatorname{card} \biggl\{ m\geq0:0< \frac {(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}=\alpha \biggr\} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \biggl\{ m\geq0:0< \frac{(4mk+4k-2i-1)\pi }{4k} \biggl|\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \biggr|=\alpha \biggr\} , \quad \alpha\geq0, \\ N^{0}(\beta_{-}) =&\sum_{i=0}^{k_{1}} \operatorname{card} \biggl\{ m\geq0:0< \frac {(4mk+4k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}=-\beta \biggr\} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \biggl\{ m\geq0:0< \frac{(4mk+2i+1)\pi }{4k} \biggl|\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \biggr|=-\beta \biggr\} , \quad \beta< 0, \\ N^{0}(\beta_{+}) =&\sum_{i=0}^{k_{1}} \operatorname{card} \biggl\{ m\geq0:0< \frac {(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}=\beta \biggr\} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \biggl\{ m\geq0:0< \frac{(4mk+4k-2i-1)\pi }{4k} \biggl|\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \biggr|=\beta \biggr\} ,\quad \beta\geq0. \end{aligned}

Now we give the main results of this paper.

### Theorem 4.1

Suppose that (S1) and (S2) hold. Then equation (1.6) possesses at least

$$n=\max \bigl\{ N(\beta)-N(\alpha)-N^{0}(\beta_{-})-N^{0}( \alpha_{+}), N(\alpha)-N(\beta)-N^{0}(\alpha_{-})-N^{0}( \beta_{+}) \bigr\}$$

4k-periodic solutions satisfying $$x(t-2k)=-x(t)$$, provided that $$n > 0$$.

### Theorem 4.2

Suppose that (S1), (S2), ($$\mathrm{S}_{3}^{+}$$), and ($$\mathrm{S}_{4}^{-}$$) hold. Then equation (1.6) possesses at least

$$n=N(\beta)-N(\alpha)+N^{0}(\beta_{+})+N^{0}( \alpha_{-})$$

4k-periodic solutions satisfying $$x(t-2k)=-x(t)$$, provided that $$n > 0$$.

### Theorem 4.3

Suppose that (S1), (S2), ($$\mathrm{S}_{3}^{-}$$), and ($$\mathrm{S}_{4}^{+}$$) hold. Then equation (1.6) possesses at least

$$n=N(\alpha)-N(\beta)+N^{0}(\alpha_{+})+N^{0}( \beta_{-})$$

4k-periodic solutions satisfying $$x(t-2k)=-x(t)$$, provided that $$n > 0$$.

## 5 Proof of main results of this paper

### Proof of Theorem 4.1

Suppose without loss of generality that

$$n=N(\beta)-N(\alpha)-N^{0}(\beta_{-})-N^{0}( \alpha_{+}).$$

Let $$X^{+}=X_{\infty}^{+}$$ and $$X^{-}=X_{0}^{-}$$. Then

$$X\setminus \bigl(X^{+}\cup X^{-} \bigr)=X\setminus \bigl(X_{\infty}^{+}\cup X_{0}^{-} \bigr) \subseteq X_{\infty}^{0}\cup X_{0}^{0}\cup \bigl(X_{\infty }^{+}\cap X_{0}^{-} \bigr).$$

Obviously,

$$\operatorname{co\,dim}_{X} \bigl(X^{+}+X^{-} \bigr)\leq\dim X_{\infty}^{0}+\dim X_{0}^{0}+ \dim \bigl(X_{\infty}^{+}\cap X_{0}^{-} \bigr)< \infty,$$

which implies that condition (a) in Lemma 1.1 holds. Let $$A_{\infty}=\beta$$. Then condition (b) in Lemma 1.1 holds since for each $$j\in N$$, we have that $$x\in X(j)$$ yields $$(L+P^{-1}\beta)x\in X(j)$$.

At the same time, Lemma 3.2 gives the $$(P.S)$$-condition.

Now it suffices to show that conditions (c) and (d) in Lemma 1.1 hold under assumptions (S1) and (S2).

In fact, condition (S1) implies that on $$X^{-}$$ we have $$\Phi (x)>0$$ if $$0<\|x\|\ll1$$, that is, there are $$r>0$$ and $$c_{\infty}<0$$ such that

$$\Phi(x)\leq c_{\infty}< 0=\Phi(0),\quad\forall x\in X^{-}\cap S_{r}=\bigl\{ x\in X:\|x\|=r\bigr\} .$$

On the other hand, we have shown in Lemma 3.1 that there is $$\sigma>0$$ such that $$\langle(L+P^{-1})x,x \rangle>\sigma\|x\|^{2},x\in X_{\infty}^{+}$$. On the other hand, $$|F(x)-\frac{1}{2}\beta x^{2}|<\frac{1}{4}\sigma|x|^{2}+M_{1},x\in R$$, for some $$M_{1}>0$$.

Then

\begin{aligned} \Phi(x) =&\frac{1}{2} \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle + \int _{0}^{4k} \biggl[F \bigl(x(t) \bigr)- \frac{1}{2}\beta\bigl|x(t)\bigr|^{2} \biggr]\,dt \\ \geq&\frac{1}{2}\sigma\|x\|^{2}-\frac{1}{4}\sigma\|x \|^{2}-4kM_{1} \\ \geq&\frac{1}{4}\sigma\|x\|^{2}-4kM_{1} \end{aligned}

if $$x\in X^{+}$$. Clearly, there is $$c_{0}< c_{\infty}$$ such that $$\Phi (x)\geq c_{0},x\in X^{+}$$.

Our last task is to compute the value of

\begin{aligned} n =&\frac{1}{2} \bigl[\dim \bigl(X^{+}\cap X^{-} \bigr)-\operatorname{co\,dim}_{X} \bigl(X^{+}+X^{-} \bigr) \bigr] \\ =&\frac{1}{2} \bigl[\dim \bigl(X_{\infty}^{+}\cap X_{0}^{-} \bigr)-\operatorname{co\,dim} _{X} \bigl(X_{\infty}^{+}+X_{0}^{-} \bigr) \bigr] \\ =&\frac{1}{2}\sum_{j=0}^{\infty} \bigl[ \dim \bigl(X_{\infty}^{+}(j)\cap X_{0}^{-}(j) \bigr)-\operatorname{co\,dim}_{X(j)} \bigl(X_{\infty}^{+}(j)+X_{0}^{-}(j) \bigr) \bigr]. \end{aligned}

By computation we get that, for each $$i\in\{0,1,\ldots,k_{2}\}$$,

\begin{aligned}& \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle = \biggl(- \frac{\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\frac{\beta}{4mk+2i+1} \biggr)\|x \|^{2},\quad x\in X(2mk+i), \\& \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle = \biggl( \frac{\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\frac{\beta}{4mk+4k-2i-1} \biggr)\|x\| ^{2},\\& \quad x\in X(2mk+2k-i-1), \end{aligned}

and

\begin{aligned}& \bigl\langle \bigl(L+P^{-1}\alpha \bigr)x,x \bigr\rangle = \biggl(- \frac{\pi}{4k}\frac {\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\frac{\alpha}{4mk+2i+1} \biggr)\|x\| ^{2},\quad x\in X(2mk+i), \\& \bigl\langle \bigl(L+P^{-1}\alpha \bigr)x,x \bigr\rangle = \biggl( \frac{\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\frac{\alpha}{4mk+4k-2i-1} \biggr)\|x\| ^{2}, \\& \quad x\in X(2mk+2k-i-1). \end{aligned}

Therefore,

\begin{aligned}& X_{\infty}^{+}(2mk+i)=X_{\infty}^{+}\cap X(2mk+i)=\emptyset, \\& X_{\infty}^{+}(2mk+2k-i-1)=X_{\infty}^{+}\cap X(2mk+2k-i-1)=X(2mk+2k-i-1), \\& X_{0}^{-}(2mk+i)=X_{0}^{-}\cap X(2mk+i)=X(2mk+i), \\& X_{0}^{-}(2mk+2k-i-1)=X_{0}^{-}\cap X(2mk+2k-i-1)=\emptyset \end{aligned}

if $$i\in\{0,1,\ldots,k_{1}\}$$ and $$m\geq0$$ is large enough,

\begin{aligned}& X_{\infty}^{+}(2mk+i)=X_{\infty}^{+}\cap X(2mk+i)=X(2mk+i), \\& X_{\infty}^{+}(2mk+2k-i-1)=X_{\infty}^{+}\cap X(2mk+2k-i-1)=\emptyset, \\& X_{0}^{-}(2mk+i)=X_{0}^{-}\cap X(2mk+i)= \emptyset, \\& X_{0}^{-}(2mk+2k-i-1)=X_{0}^{-}\cap X(2mk+2k-i-1)=X(2mk+2k-i-1) \end{aligned}

if $$i\in\{k_{1}+1,\ldots,k_{2}\}$$ and $$m\geq0$$ is large enough, which means that there is $$M>0$$ such that $$\dim(X_{\infty}^{+}(j)\cap X_{0}^{-}(j))-\operatorname{co\,dim}_{X}(X_{\infty }^{+}(j)+X_{0}^{-}(j))=0, j>M$$, from which it follows that

\begin{aligned} n =&\frac{1}{2}\sum_{j=0}^{M} \bigl[ \dim \bigl(X_{\infty}^{+}(j)\cap X_{0}^{-}(j) \bigr)-\operatorname{co\,dim}_{X(j)} \bigl(X_{\infty}^{+}(j)+X_{0}^{-}(j) \bigr) \bigr] \\ =&\frac{1}{2}\sum_{j=0}^{M} \bigl[ \dim X_{\infty}^{+}(j)+\dim X_{0}^{-}(j)-2 \bigr] \\ =&\frac{1}{2}\sum_{j=0}^{M} \bigl[ \dim X_{\infty}^{+}(j)+\dim X_{0}^{-}(j) \bigr]-(M+1). \end{aligned}

Then we have

\begin{aligned}& \sum_{j=0}^{M}\dim \bigl(X_{\infty}^{+}(j) \bigr) \\& \quad=2\left \{ \textstyle\begin{array}{@{}l} N(\beta)+\operatorname{card}\{2mk+2k-i-1:0\leq2mk+2k-i-1\leq M,0\leq i\leq k_{1}\}\\ \quad{}+\operatorname{card}\{2mk+i:0\leq2mk+i\leq M,k_{1}+1\leq i\leq k_{2}\}, \quad\beta>0,\\ N(\beta)-N^{0}(\beta_{-})+\operatorname{card}\{2mk+2k-i-1:0\leq2mk+2k-i-1\leq M,0\leq i\leq k_{1}\}\\ \quad{}+\operatorname{card}\{2mk+i:0\leq2mk+i\leq M,k_{1}+1\leq i\leq k_{2}\}, \quad\beta< 0, \end{array}\displaystyle \right . \\& \sum_{j=0}^{M}\dim \bigl(X_{0}^{-}(j) \bigr) \\& \quad=2\left \{ \textstyle\begin{array}{@{}l} -N(\alpha)-N^{0}(\alpha_{+})+\operatorname{card}\{2mk+i:0\leq2mk+i\leq M,0\leq i\leq k_{1}\}\\ \qquad{}+\operatorname{card}\{2mk+2k-i-1:0\leq2mk+2k-i-1\leq M,k_{1}+1\leq i\leq k_{2}\},\\ \quad\alpha>0,\\ -N(\alpha)+\operatorname{card}\{2mk+i:0\leq2mk+i\leq M,0\leq i\leq k_{1}\}\\ \qquad{}+\operatorname{card}\{2mk+2k-i-1:0\leq2mk+2k-i-1\leq M,k_{1}+1\leq i\leq k_{2}\},\\ \quad\alpha< 0, \end{array}\displaystyle \right . \end{aligned}

and

$$\sum_{j=0}^{M} \bigl[\dim X_{\infty}^{+}(j)+\dim X_{0}^{-}(j) \bigr]= 2 \bigl[N(\beta)-N(\alpha)-N^{0}(\beta_{-})-N^{0}( \alpha_{+}) \bigr]+2(M+1).$$
(5.1)

Therefore,

$$n=N(\beta)-N(\alpha)-N^{0}(\beta_{-})-N^{0}( \alpha_{+}) .$$

Theorem 4.1 is proved. □

### Proof of Theorem 4.2 and Theorem 4.3

Since the proof for the two theorems is almost the same, we prove only Theorem 4.2.

Let $$X^{+}=X_{\infty}^{+}+X_{\infty}^{0},X^{-}=X_{-}^{0}+X_{0}^{0}$$. Then as in the proof of Theorem 4.1, we check conditions (a), (b), (c), (d), and (e). In the present case, we may suppose that (5.1) still holds for some $$M>0$$. Let $$X_{\infty}^{0}(i)=X_{\infty}^{0}\cap X(i),X_{0}^{0}(i)=X_{0}^{0}\cap X(i)$$. Then

\begin{aligned} n =&\frac{1}{2}\sum_{i=0}^{M} \bigl[ \dim \bigl(X_{\infty}^{+}(i)\cap X_{0}^{-}(i) \bigr)-\operatorname{co\,dim}_{X(i)} \bigl(X_{\infty}^{+}(i)+X_{0}^{-}(i) \bigr) \bigr]+ \bigl(\dim X_{\infty}^{0}+\dim X_{0}^{0} \bigr) \\ =&\frac{1}{2}\sum_{i=0}^{M} \bigl[ \dim X_{\infty}^{+}(i)+\dim X_{0}^{-}(i)-2 \bigr]+ \bigl(\dim X_{\infty}^{0}+\dim X_{0}^{0} \bigr) \\ =&\frac{1}{2}\sum_{i=0}^{M} \bigl[ \dim X_{\infty}^{+}(i)+\dim X_{0}^{-}(i) \bigr]-(M+1)+ \bigl(\dim X_{\infty}^{0}+\dim X_{0}^{0} \bigr) \\ =&N(\beta)-N(\alpha)-N^{0}(\beta_{-})-N^{0}( \alpha_{+})+ \bigl(N^{0}(\beta_{+})+N^{0}( \beta_{-})+N^{0}(\alpha_{+})+N^{0}(\alpha _{-}) \bigr) \\ =&N(\beta)-N(\alpha)+N^{0}(\beta_{+})+N^{0}( \alpha_{-}). \end{aligned}

Our proof is completed. □

## 6 An example

### Example

Suppose that $$f\in C^{0}(R,R)$$ is such that

$$f(x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{5\pi}{2}x+x^{\frac{1}{3}}, &|x|\gg1,\\ -\frac{7\pi}{2}x-x^{3}, & |x|\ll1. \end{array}\displaystyle \right .$$

We are to discuss the multiplicity of 12-periodic solutions of the equation

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x(t-5) \bigr).$$
(6.1)

In this case, $$k=3,k_{1}=1,k_{2}=2,\alpha=-\frac{7\pi}{2},\beta=\frac {5\pi}{2}$$. This yields that

\begin{aligned}& \begin{aligned}[b] N(\alpha)={}&{-}\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+12-1)\pi}{12} \frac{\cos\frac{\pi }{12}}{\sin\frac{3\pi}{12}}< \frac{7\pi}{2} \biggr\} \\ &{}-\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+12-3)\pi}{12}\frac{\cos\frac{3\pi }{12}}{\sin\frac{9\pi}{12}}< \frac{7\pi}{2} \biggr\} \\ &{}-\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+5)\pi}{12} \biggl|\frac{\cos\frac{5\pi }{12}}{\sin\frac{15\pi}{12}} \biggr|< \frac{7\pi}{2} \biggr\} =-15, \end{aligned}\\& \begin{aligned}[b] N(\beta)={}&\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+1)\pi}{12} \frac{\cos\frac{\pi }{12}}{\sin\frac{3\pi}{12}}< \frac{5\pi}{2} \biggr\} \\ &{}+\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+3)\pi}{12}\frac{\cos\frac{3\pi}{12}}{\sin \frac{9\pi}{12}}< \frac{5\pi}{2} \biggr\} \\ &{}+\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+12-5)\pi}{12} \biggl|\frac{\cos\frac{5\pi }{12}}{\sin\frac{15\pi}{12}} \biggr|< \frac{5\pi}{2} \biggr\} =12, \end{aligned}\\& N^{0}(\alpha_{+})=N^{0}(\beta_{-})=N^{0}( \alpha_{-})=N^{0}(\beta_{+})=0. \end{aligned}

Applying Theorem 4.2, we conclude that equation (6.1) possesses at least 27 different 12-periodic orbits satisfying $$x(t-6)=-x(t)$$ since f satisfies (S1), (S2), ($$\mathrm{S}_{3}^{+}$$), and ($$\mathrm{S}_{4}^{-}$$).

## References

1. Briat, C: Linear Parameter-Varying and Time-Delay Systems: Analysis, Observation, Filtering and Control. Springer, Berlin (2015)

2. Nesbit, R: Delay differential equations for structured populations. In: Structured Population Models in Marine, Terrestrial and Freshwater Systems (1997)

3. Wang, DS, Hu, XH, Hu, J, Liu, WM: Quantized quasi-two-dimensional Bose-Einstein condensates with spatially modulated nonlinearity. Phys. Rev. A 81, 025604 (2010)

4. Wang, DS, Yin, SJ, Tian, Y, Liu, Y: Integrability and bright soliton solutions to the coupled nonlinear Schrödinger equation with higher-order effects. Appl. Math. Comput. 229, 296-309 (2014)

5. Wang, DS, Wei, X: Integrability and exact solutions of a two-component Korteweg-de Vries system. Appl. Math. Lett. 51, 60-67 (2016)

6. Kaplan, J, Yorke, J: Ordinary differential equations which yield periodic solution of delay equations. J. Math. Anal. Appl. 48, 317-324 (1974)

7. Nussbaum, R: Periodic solutions of special differential delay equations: an example in nonlinear functional analysis. Proc. R. Soc. Edinb. 81(1-2), 131-151 (1977)

8. Fei, G: Multiple periodic solutions of differential delay equations via Hamiltonian systems (I). Nonlinear Anal. 65, 25-39 (2006)

9. Fei, G: Multiple periodic solutions of differential delay equations via Hamiltonian systems (II). Nonlinear Anal. 65, 40-58 (2006)

10. Ge, W: Periodic solutions of the differential delay equation $$x'(t)=-f(x(t-1))$$. Acta Math. Sin. New Ser. 12, 113-121 (1996)

11. Ge, W: On the existence of periodic solutions of differential delay equations with multiple lags. Acta Math. Appl. Sin. 17(2), 173-181 (1994) (in Chinese)

12. Ge, W: Two existence theorems of periodic solutions for differential delay equations. Chin. Ann. Math., Ser. B 15(2), 217-224 (1994)

13. Ge, W: Oscillatory periodic solutions of differential delay equations with multiple lags. Chin. Sci. Bull. 42(6), 444-447 (1997)

14. Ge, W, Zhang, L: Multiple periodic solutions of delay differential systems with $$2k-1$$ lags via variational approach. Discrete Contin. Dyn. Syst. 36(9), 4925-4943 (2016)

15. Ge, W, Zhang, L: Multiple periodic solutions of delay differential systems with 2k lags via variational approach. Preprint

16. Guo, Z, Yu, J: Multiple results for periodic solutions to delay differential equations via critical point theory. J. Differ. Equ. 218, 15-35 (2005)

17. Guo, Z, Yu, J: Multiple results on periodic solutions to higher dimensional differential equations with multiple delays. J. Dyn. Differ. Equ. 23, 1029-1052 (2011)

18. Li, J, He, X: Multiple periodic solutions of differential delay equations created by asymptotically linear Hamiltonian systems. Nonlinear Anal. TMA 31, 45-54 (1998)

19. Li, J, He, X: Proof and generalization of Kaplan-Yorke’ conjecture under the condition $$f'(0)>0$$ on periodic solution of differential delay equations. Sci. China Ser. A 42(9), 957-964 (1999)

20. Li, J, He, X, Liu, Z: Hamiltonian symmetric group and multiple periodic solutions of differential delay equations. Nonlinear Anal. TMA 35, 957-964 (1999)

21. Li, S, Liu, J: Morse theory and asymptotically linear Hamiltonian systems. J. Differ. Equ. 78, 53-73 (1989)

22. Zhang, L, Ge, W: Two classes of periodic solutions to Kaplan-Yorke like differential delay equations with two lags of integer ratio. Preprint

23. Benci, V: On critical point theory for indefinite functionals in the presence of symmetries. Trans. Am. Math. Soc. 274(2), 533-572 (1982)

24. Fannio, L: Multiple periodic solutions of Hamiltonian systems with strong resonance at infinity. Discrete Contin. Dyn. Syst. 3, 251-264 (1997)

25. Mawhen, J, Willem, M: Critical Point Theory and Hamiltonian Systems. Springer, New York (1989)

## Acknowledgements

Sponsored by the Scientific Research Project of Beijing Municipal Commission of Education (No. KM201511232018) and the National Natural Science Foundation of China (11471146).

## Author information

Authors

### Corresponding author

Correspondence to Chunyan Xue.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

## Rights and permissions

Reprints and permissions

Li, L., Xue, C. & Ge, W. Periodic orbits to Kaplan-Yorke like differential delay equations with two lags of ratio $$(2k-1)/2$$ . Adv Differ Equ 2016, 247 (2016). https://doi.org/10.1186/s13662-016-0967-3